In a test an examiner either guesses or copies or knows the answer...

Let A$be the event that the examinee gives the correct answer.</p><p>Let$G, C, K$stand for guessing, copying and knowing. respectively.</p><p>Given,$P\left(C\right) = \dfrac{1}{6}, P\left(G\right) = \dfrac{1}{3}$</p><p>and$P\left(\dfrac{A}{C}\right) = \dfrac{1}{8}$</p><p>$⇒ P\left(C\right) + P\left(G\right) + P\left(K\right) = 1$</p><p>$⇒ \dfrac{1}{6} + \dfrac{1}{3} + P\left(K\right) = 1$</p><p>$∴ P\left(K\right) = \dfrac{1}{2}$</p><p>Also,$P\left(\dfrac{A}{K}\right) = 1,$for if the examinee knows, he/she will correctly answer it and$P\left(\dfrac{A}{G}\right) = \dfrac{1}{4},</p><p> since there are four choices.</p><p>Now, total probability&P\left(A\right) = P\left(G\right)P\left(\dfrac{A}{G}\right) + P \left(C\right)P \left(\dfrac{A}{C}\right) + +P\left(K\right)P\left(\dfrac{A}{K}\right)$</p><p>$= \dfrac{1}{3} \cdot \dfrac{1}{4} + \dfrac{1}{6} \cdot \dfrac{1}{8} + \dfrac{1}{2} \cdot1$</p><p>$= \dfrac{1}{12} + \dfrac{1}{48} + \dfrac{1}{2} = \dfrac{4 + 1 + 24}{48} = \dfrac{29}{48}$</p><p>Now,$P\left(\dfrac{K}{A}\right) = \dfrac{P\left(K\right)P\left(\dfrac{A}{K}\right)}{P\left(A\right)}$</p><p>$= \dfrac{\dfrac{1}{2} \cdot 1}{2948} = \dfrac{1}{2} \cdot \dfrac{48}{29} = \dfrac{24}{29}$